Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<iostream>
#include<map>
using namespace std;

typedef long long ll;
const int mo = 998244353, X = 1e4 + 7, Y = 1e7 + 7;
const int N = 2.5e5 + 10;
int n;
ll cnt[N], pre[N], X1, cnt2[N], pre2[N], Y1;
char s[N];

ll ksm(ll x, ll y) {
	ll ret = 1; for (; y; y >>= 1) {
		if (y & 1) ret = ret * x % mo;
		x = x * x % mo;
	}
	return ret;
}

map<pair<ll, ll>, int> h;
pair<ll, ll> tj(int x) {
	return make_pair(((pre[n] * cnt[x] % mo + pre[x]) % mo + mo) % mo,
		((pre2[n] * cnt2[x] % mo + pre2[x]) % mo + mo) % mo);
}

int main() {
	cin >> n;
	scanf("%s", s + 1);
	cnt[0] = 1; cnt2[0] = 1;
	X1 = ksm(X, mo - 2); Y1 = ksm(Y, mo - 2);
	for (int i = 1; i <= n; i++) {
		cnt[i] = cnt[i - 1];
		pre[i] = pre[i - 1];
		cnt2[i] = cnt2[i - 1];
		pre2[i] = pre2[i - 1];
		if (s[i] == '>') {
			cnt[i] = cnt[i] * X % mo;
			cnt2[i] = cnt2[i] * Y % mo;
		}
		else if (s[i] == '<') {
			cnt[i] = cnt[i] * X1 % mo;
			cnt2[i] = cnt2[i] * Y1 % mo;
		}
		else if (s[i] == '+') {
			pre[i] = (pre[i] + cnt[i]) % mo;
			pre2[i] = (pre2[i] + cnt2[i]) % mo;
		}
		else {
			pre[i] = (pre[i] - cnt[i]) % mo;
			pre2[i] = (pre2[i] - cnt2[i]) % mo;
		}
	}
	ll ans = 0;
	h[tj(0)] = 1;
	for (int i = 1; i <= n; i++) {
		pre[i] = (pre[i] + mo) % mo;
		pre2[i] = (pre2[i] + mo) % mo;
		ans += h[make_pair(pre[i], pre2[i])];
		h[tj(i)] ++;
	}
	cout << ans << endl;

	return 0;
}